Inverse dynamic analyzing of flexible link manipulators with translational and rotational joints
Nonlinear dynamic modeling and equations of
motion of flexible manipulators with translational
and rotational joints are built by using finite
element method and Lagrange approach. Model is
developed based on single link manipulator with
only rotational joint. Inverse dynamic problem of
flexible link manipulator is surveyed with an
algorithm which is based on rigid model.
Approximate driving force and torque at joints of
flexible link manipulator are found with desire
path. Derivation values of these also are shown.
Elastic displacements at end-effector point are
presented. However, there are remaining issues
which need be studied further in future work
because the error joints variables in algorithm to
solve inverse dynamic problem of flexible with
translational joint has not been mentioned yet.
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42 SCIENCE & TECHNOLOGY DEVELOPMENT, Vol 20, No.K2- 2017
Abstract— Inverse dynamic problem analyzing of
flexible link robot with translational and rotational
joints is presented in this work. The new model is
developed from single flexible link manipulator with
only rotational joint. The dynamic equations are built
by using finite element method and Lagrange
approach. The approximate force of translational
joint and torque of rotational joint are found based
on rigid model. The simulation results show the
values of driving forces at joints of flexible robot with
desire path and errors of joint variables between
flexible and rigid models. Elastic displacements of
end-effector are shown, respectively. There are
remaining issues which need be studied further in
future work because the error joints variables in
algorithm to solve inverse dynamic problem of
flexible with translational joint has not been
mentioned yet.
Index Terms—Inverse Dynamic , flexible link
manipulator, translational joint, elastic
displacements.
1 INTRODUCTION
ynamic analysis of mechanisms, especially
robots, is very important. The dynamic
equations of motion represent the behavior
of system, so accurate modeling and equations are
essential to successfully design of the control
system. The analysis of robots considering the
elastic characteristics of its members has been
considerable attention in recent years. Flexibility in
robots can affect position accuracy. Inverse
Manuscript Received on July 13th, 2016. Manuscript Revised
December 06th, 2016.
Bien Xuan Duong, My Anh Chu are with Military Technical
Academy. Email: xuanbien82@yahoo.com.
Khoi Bui Phan is with Ha Noi University of Science and
Technology, Ha Noi.
dynamic of flexible robots is very essential for
selecting the actuator and designing the proper
control strategy. Most of the investigations on the
dynamic modeling of robot manipulators with
elastic arms have been confined to manipulators
with only revolute joints.
In the literature, most of the investigations on the
inverse dynamics of the flexible robot manipulator
copies with manipulators constructed with only
rotational joints [1-3]. Kwon and Book [1] present
a single link robot which is described and modeled
by using assumed modes method (AMM). Inverse
equation is derived in a state space form from direct
dynamic equations and using definitions concepts
which are causal system, anti-causal system and
Non-causal system. Based on these concepts, the
time-domain inverse dynamic method was
interpreted in the frequency-domain in detail by
using the two sided Laplace transform in the
frequency-domain and the convolution integral.
This method is limited to linear system. Stable
inversion method is studied for the same robot
configuration but the nonlinear effect is taken into
account [2]. An inversion-based approach to exact
nonlinear output tracking control is presented. Non-
causal inversion is incorporated into tracking
regulators and is a powerful tool for control.
Eliodoro and Miguel [3] propose a new method
based on the finite difference approach to discretize
the time variable for solving the inverse dynamics
of the robot. This method is a non-recursive and
non-iterative approach carried out in the time
domain in contrast with methods previously
proposed. By using either the finite element method
(FEM) or AMM, some other authors consider the
dynamic modeling and analysis of the flexible
robots with translational joint [4-8]. Pan et al [4]
presented a model R-P with FEM approach. The
Inverse dynamic analyzing of flexible link
manipulators with translational and rotational
joints
Bien Xuan Duong, My Anh Chu, and Khoi Bui Phan
D
TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ K2-2017
43
result is differential algebraic equations which are
solved by using Newmark method. Al-Bedoor and
Khulief [5] presented a general dynamic model for
R-P robot based on FEM and Lagrange approach.
They defined a concept which is translational
element. The stiffness of translational element is
changed. The prismatic joint variable is distance
from origin coordinate system to translational
element. The number of element is small because it
is hard challenge to build and solve differential
equations. Khadem [6] studied a three-dimensional
flexible n-degree of freedom manipulator having
both revolute and prismatic joint. A novel approach
is presented using the perturbation method. The
dynamic equations are derived using the Jourdain’s
principle and the Gibbs-Appell notation. Korayem
[7] also presented a systematic algorithm capable of
deriving equations of motion of N-flexible link
manipulators with revolute-prismatic joints by
using recursive Gibbs-Appell formulation and
AMM. However, the inverse dynamics modeling
and analysis of the generalized flexible robot
constructed with translational joint has not been
much mentioned yet.
The objective of the described work in what
follows was to present surveying inverse dynamics
problem of flexible link robot with translational and
rotational joints. The Lagrange approach in
conjunction with the finite element method is
employed in deriving the equations of motion.
Inverse dynamics problem of model with flexible
link can be approximately solved based on model
with rigid links. The forward kinematic, inverse
kinematic and inverse dynamics of rigid model are
used to find joints values from desire path and
driving force and torque which are inputs data for
flexible model problems. The force and torque of
joints can be found in such a way that the end point
of link 2 can track the desire path even though link
2 is deformed.
2 DYNAMIC MODELING
2.1 Dynamic model
In this work, we concern the dynamic model of
two link flexible robot which motions on horizontal
plane with translational joint for first rigid link and
rotational joint for second flexible link to formulate
the inverse dynamics problem. It is shown as Fig 1.
Figure 1. Flexible links robot with translational and rotational
joints
The coordinate system XOY is the fixed frame.
Coordinate system 1 1 1X O Y is attached to end point
of link 1. Coordinate system 2 2 2X O Y is attached to
first point of link 2. The translational joint variable
d t is driven by TF t force. The rotational joint
variable q t
is driven by t torque. Both joints
are assumed rigid. Link 1 with length 1L
is
assumed rigid and link 2 with length 2L
is assumed
flexibility. Link 2 is divided n elements. The
elements are assumed interconnected at certain
points, known as nodes. Each element has two
nodes. Each node of element j has 2 elastic
displacement variables which are the flexural
2 1 2 1,j ju u- and the slope displacements
2 2 2,j ju u . Symbol tm is the mass of payload on
the end-effector point. The coordinate 01r of end
point of link 1 on XOY is computed as
01 1
T
L d t= r (1)
The coordinate 2 jr of element j on 2 2 2X O Y can
be given as
2 j1 , ; 0..
T
j e j j j ej l x w x t x l = - = r (2)
Where, length of each element is 2e
L
l
n
= and
,j jw x t is the total elastic displacement of
element j which is defined by [10]
,j j j j jw x t x t= N Q (3)
The vector of shape function j jxN is defined
as
44 SCIENCE & TECHNOLOGY DEVELOPMENT, Vol 20, No.K2- 2017
1 2 3 4j j j j j jx x x x x = N f f f f (4)
Mode shape function ; ( 1...4)i jx i =f can be
presented in [10]. The elastic displacement
j
tQ
of element j is given as
2 1 2 2 1 2 2
T
j j j jj
t u u u u- = Q (5)
The coordinate 21 jr of element j on 1 1 1X O Y can
be written as
1
21 2 2.j j=r T r (6)
Where,
1
2
cos sin
sin cos
q t q t
q t q t
-
=
T is the
transformation matrix from 2 2 2X O Y to 1 1 1X O Y .The
coordinate 02 jr of element j on XOY can be
computed as
02 1 21j j= r r r (7)
The elastic displacement
n
tQ of element n is
given as
2 1 2 2 1 2 2
T
n n n nn
t u u u u- =Q (8)
The coordinate 0Er of end point of flexible link 2
on XOY can be computed as
1 2 2 1
0
2 2 1
cos sin
sin cos
n
E
n
L L q t u q t
d t L q t u q t
-
=
r (9)
If assumed that robot with all of links are rigid,
the coordinate 0 _E rigidr on XOY can be written as
1 2
0 _
2
cos
sin
E rigid
L L q t
d t L q t
=
r (10)
The kinematic energy of link 1 can be computed
as
2
1 1 01
1
.
2
T m= r (11)
Where 1m is the mass of link 1. The kinetic
energy of element j is determined as
2
02
2 20
1 1
2 2
el j T
j j jg j jgT m dx t t
t
= =
r
Q M Q (12)
Where 2m is mass per meter of link 2. The
generalized elastic displacement
jg
tQ of
element j is given as
2 1 2 2 1 2 2
T
j j j jjg
t d t q t u u u u- = Q
(13)
Each element of inertial mass matrix jM can be
computed as
02 0220, ; , 1, 2,.., 6
= =
e
T
l j j
j j
js je
s e m dx s e
Q Q
r r
M (14)
Where jsQ and jeQ are the ,
th ths e element of jgQ
vector. It can be shown that jM
is of the form
11 12 13 14 15 16
21 22 23 24 25 26
31 32
41 42 _
51 52
61 62
=
j
j base
m m m m m m
m m m m m m
m m
m m
m m
m m
M
M
(15)
Where,
2 2
2 2 2 2
2 3 2 3
2 2 2 2
_
2 2
2 2 2 2
2 3 2 3
2 2 2 2
13 11 9 13
35 210 70 420
11 1 13 1
210 105 420 140
9 13 13 11
70 420 35 210
13 1 11 1
420 140 210 105
-
-
=
-
- - -
e e e e
e e e e
j base
e e e e
e e e e
m l m l m l m l
m l m l m l m l
m l m l m l m l
m l m l m l m l
M
(16)
And,
2 1 2 1 2
11 2 12 2
2 2
2
13 15 2 14 16 2 21 12
2 3 2
23 2 24 2 25 2
22 2
(6 61
. ; . ;
)sin 6 (1 2 ) cos12
1 1
cos ; cos ; ;
2 2
1 1 1
(10 7); (5 3); (10 3);
20 60 20
210
1
210
j j e j
e e
e j e
e e
e e e
e
u u l u
m m l m m l
l u q l j q
m m m l q m m m l q m m
m m l j m m l j m m l j
m m l
-
= = -
- -
= = = = =
= - = - = -
=
2 2 2 2
2 2 2 2 2 2
2 1 2 2 1 2 2 2 2 1 2 1 2 2
2 2 2
2 1 2 1 2 1 2 1
3
26 2 31 13 32 23 41 14 42 24
51 15
( 1) (2 3 2 )
22 ( ) 13 ( ) ;
78( ) 70 54
1
(5 2); ; ; ; ;
60
;
e e j j j j
e j j j j e j j j j
j j e j j
e
l j j l u u u u
l u u u u l u u u u
u u l u u
m m l j m m m m m m m m
m m
- -
- -
- -
- -
= - - = = = =
= 52 25 61 16 62 26; ;m m m m m m= = =
The total elastic kinetic energy of link 2 is yielded
as
2
1
1
2
n
T
dh j dh
j
T T t t
=
= = Q M Q (17)
The inertial mass matrix dhM is constituted from
matrices of elements follow FEM theory,
respectively. Vector tQ represents the
generalized coordinate of system and is given as
1 2 1 2 2. .
T
n nt d t q t u u u = Q (18)
The kinetic energy of payload is given as
TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ K2-2017
45
2
0
1
.
2
P t ET m= r (19)
The kinetic energy of system is determined as
1
1
2
T
dh PT T T T t t== + + Q MQ
(20)
Matrix M is mass matrix of system. The gravity
effects can be ignored as the robot movement is
confined to the horizontal plane. Defining E and
I are Young’s modulus and inertial moment of
link 2, the elastic potential energy of element j is
shown as jP
with the stiffness matrix jK and
presented as [10]
2
2
20
,1 1
2 2
el j j T
j j j j j
j
w x t
P EI dx t t
x
= =
Q K Q (21)
Where,
3 2 3 2
2 2
3 2 3 2
2 2
0 0 0 0 0 0
0 0 0 0 0 0
12 6 12 6
0 0
6 4 6 2
0 0
12 6 12 6
0 0
6 2 6 4
0 0
-
-=
- - -
-
e e e e
j
e ee e
e e e e
e ee e
EI EI EI EI
l l l l
EI EI EI EI
l ll l
EI EI EI EI
l l l l
EI EI EI EI
l ll l
K
(22)
The total elastic potential energy of system is
yielded as
1
1
2
n
T
j
j
P P t t
=
= = Q KQ (23)
The stiffness matrix K is constituted from
matrices of elements follow FEM theory similar
M matrix, respectively.
2.2 Dynamic equations of motion
Fundamentally, the method relies on the
Lagrange equations with Lagrange function
L T P= - are given by
( )
d L L
t
dt tt
- =
F
QQ
(24)
Vector tF is the external generalized forces
acting on specific generalized coordinate tQ and
is determined as
0 . . 0 0
T
Tt F t t= F (25)
Size of matrices ,M K is 2 4 2 4n n and
size of tF and tQ is 2 4 1n . The
rotational joint of link 2 is constrained so that the
elastic displacements of first node of element 1 on
link 2 can be zero. Thus variables 1 2,u u are zero.
By enforcing these boundary conditions and FEM
theory, the generalized coordinate tQ becomes
3 2 1 2 2. .
T
n nt d t q t u u u = Q (26)
So now, size of matrices ,M K is
2 2 2 2n n and size of tF and tQ is
2 2 1n . When kinetic and potential energy are
known, it is possible to express Lagrange equations
as shown
tM Q Q + C Q,Q Q + DQ + KQ = F (27)
Where structural damping D and coriolis force
C matrices are calculated as
1, ( )
2
T = -
C Q Q Q M Q Q Q M Q Q
Q
(28)
= D M K (29)
Where and are the damping ratios of the
system which are determined by experience.
3 INVERSE DYNAMIC ANALYZING
Solving inverse dynamics problem can be
computed a feed-forward control to follow a
trajectory more accurately. Inverse dynamics of
flexible robot is the process of determining load
profiles to produce given displacement profiles as
function of time. Forward dynamics of flexible
robot is process of finding displacements given the
loads. This is much simpler than inverse dynamics
process because elastic displacements do not to
know before if there are not external forces which
effect on system. Unlike the rigid link, the inverse
dynamics of flexible robot is more complex
because of links deformations. We need to
determine the force and torque of actuators in such
a way that the end point of link 2 can still track the
desire path even though link 2 is deformed. Inverse
dynamics problem of model with flexible link can
be approximately solved based on model with rigid
links. Steps to solve are shown as Fig 2. The detail
of blocks in Fig 2 is presented in Fig 3, Fig 4, Fig 5
and Fig 6.
46 SCIENCE & TECHNOLOGY DEVELOPMENT, Vol 20, No.K2- 2017
Figure 2. General diagram of inverse dynamic flexible robot algorithm
Figure 3. Diagram of inverse kinematic rigid model block
Fig. 4.Diagram of inverse dynamic rigid model block
Figure 5. Diagram of forward dynamic flexible robot block
Figure 6. Diagram of inverse dynamic flexible robot block
TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ K2-2017
47
Firstly, assuming that two link is rigid. The
translational and rotational joints of rigid model are
computed from desire path by solving inverse
kinematic rigid problem [9] which is shown in Fig.
3. Then driving force and torque at joints of rigid
model are computed by solving inverse dynamic
rigid [9] (Fig. 4). Results are input data for forward
dynamic flexible model follow equation (27) and
are shown in Fig. 5. Finally, the approximates force
and torque of joints are found by solving inverse
dynamic flexible problem with inputs data which
are joints values of rigid model and elastic
displacements. It is presented by block in Fig. 6.
4 NUMERICAL SIMULATIONS
Simulation specifications of flexible model are
given by Table 1.
TABLE 1
PARAMETERS OF DYNAMIC MODEL
Property Symbol Value
Length of link 1 (m) L1 0.05
Mass of link 1 and base
(kg)
m1 1.4
Parameters of link 2
Length of link (m) L2 0.3
Width (m) b 0.02
Thickness (m) h 0.001
Number of element n 5
Cross section area (m2) A=b.h 2.10-5
Mass density (kg/m3) 7850
Mass per meter (kg/m) m=.A 0.157
Young’s modulus
(N/m2)
E 2.1010
Inertial moment of cross
section (m4)
I=b.h3/12 1.67x10-12
Damping ratios α, β 0.005;0.007
Mass of payload (g) mt 10
Desire path on
workspace in OX axis
(m)
xE
0.25-0.1sin(t-
π/2)
Desire path on
workspace in OX axis
(y)
yE 0.1sin(t)
Time simulation (s) T 2
Simulation results for inverse dynamic of
flexible robot with translational and rotational
joints are shown from Fig 7 to Fig 16. It is
noteworthy to mention that we need to find the
initial values of joints variable at t=0 when inverse
kinematic of rigid model is solved.
Figure 7. Translational joint values of rigid and flexible model
Figure 8. Rotational joint values of rigid and flexible model
Figure 9. Deviation of translational joint variables between rigid
and flexible model
Fig. 7 and fig. 8 show the values of joint
variables between rigid and flexible model.
Translational and rotational joints values are small
because of short time simulation. Fig. 9 and fig. 10
describe deviation of these values. Maximum
deviation value of translational joint is 25 mm and
rotational joint variable is 0.17 rad. These
deviations appear from effect of elastic
displacements and error of numerical method
which is used to solving problems.
48 SCIENCE & TECHNOLOGY DEVELOPMENT, Vol 20, No.K2- 2017
Figure 10. Deviation of rotational joint variables between rigid
and flexible model
Figure 11. Driving force values of rigid and flexible model
Figure 12. Driving torque values of rigid and flexible model
Figure 13. Deviation of driving force between rigid and flexible
model
Figure 14. Deviation of driving force between rigid and flexible
model
Fig. 10 to fig 14 present values of driving
forces at joints and these deviations between rigid
and flexible model. The values of driving force at
translational joint are not too difference because
first link of both models is assumed rigid.
Maximum force is 0.6 N. Driving torque values at
rotational joint are more difference because of
effect of elastic displacements of flexible link.
Figure 15. Flexural displacement value at end-effector point in
flexible model
TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ K2-2017
49
Figure 16. Slope displacement value at end-effector point in
flexible model
Fig. 15 shows flexural displacement value at
end-effect point. Maximum value is 0.7 mm. Fig.
16 shown slope displacement at end-effect point.
Maximum value is 0.035 rad. Both values are small
because of short time simulation and small values
of joint variables.
In general, simulation results show that elastic
displacements of flexible link effect on dynamic
behaviors of system. Different between rigid model
and flexible model are clearly visible.
5 CONCLUSION
Nonlinear dynamic modeling and equations of
motion of flexible manipulators with translational
and rotational joints are built by using finite
element method and Lagrange approach. Model is
developed based on single link manipulator with
only rotational joint. Inverse dynamic problem of
flexible link manipulator is surveyed with an
algorithm which is based on rigid model.
Approximate driving force and torque at joints of
flexible link manipulator are found with desire
path. Derivation values of these also are shown.
Elastic displacements at end-effector point are
presented. However, there are remaining issues
which need be studied further in future work
because the error joints variables in algorithm to
solve inverse dynamic problem of flexible with
translational joint has not been mentioned yet.
REFERENCES
[1] D. S. Kwon and W. J. Book, “A time-domain inverse
dynamic tracking control of a single link flexible
manipulator”, Journal of Dynamic Systems, Measurement
and Control, vol. 116, pp. 193–200, 1994.
[2] S. Devasia, D. Chen and B. Paden, “Non-linear inversion-
based output tracking”, IEEE Transactions on Automatic
Control, vol. 41, no. 7, 1996.
[3] C. Eliodoro and Miguel. A. Serna, “Inverse dynamics of
flexible robots”, Mathematics and computers in
simulation, 41, pp. 485-508, 1996.
[4] B. O. Al-Bedoor and Y. A. Khulief, “General planar
dynamics of a sliding flexible link”, Sound and Vibration.
206(5), pp. 641–661, 1997.
[5] Y. C. Pan, R. A. Scott, “Dynamic modeling and
simulation of flexible robots with prismatic joints”, J.
Mech. Design, 112, pp. 307–314, 1990.
[6] S. E. Khadem and A. A. Pirmohammadi, “Analytical
development of dynamic equations of motion for a three-
dimensional flexible manipulator with revolute and
prismatic joints”, IEEE Trans. Syst. Man Cybern. B
Cybern, 33(2), pp. 237–249, 2003.
[7] M. H. Korayem, A. M. Shafei and S. F. Dehkordi,
“Systematic modeling of a chain of N-flexible link
manipulators connected by revolute–prismatic joints
using recursive Gibbs-Appell formulation”, Archive of
Applied Mechanics, Volume 84, Issue 2, pp. 187–206,
2014.
[8] W. Chen, “Dynamic modeling of multi-link of flexible
robotic manipulators”, Computers and Structures, 79, pp.
183 -195, 2001.
[9] Nguyen Van Khang and Chu Anh My, Fundamentals of
Industrial Robot. Education Publisher, Ha Noi, Viet Nam,
2010, pp. 82-112.
[10] S. S. Ge, T. H. Lee and G. Zhu, “A Nonlinear feedback
controller for a single link flexible manipulator based on a
finite element method”, Journal of robotics system, 14(3),
pp. 165-178, 1997.
[11] M. O. Tokhi and A. K. M. Azad, Flexible robot
manipulators–Modeling, simulation and control.
Published by Institution of Engineering and Technology,
London, United Kingdom, 2008, pp. 113-117.
50 SCIENCE & TECHNOLOGY DEVELOPMENT, Vol 20, No.K2- 2017
Tóm tắt - Bài báo này trình bày việc phân tích bài
toán động lực học ngược của hệ rô bốt có khâu đàn
hồi với các khớp tịnh tiến và khớp quay. Mô hình
động lực học mới được phát triển từ hệ rô bốt có 1
khâu đàn hồi với chỉ một khớp quay. Hệ phương
trình động lực học được xây dựng dựa trên phương
pháp Phần tử hữu hạn và hệ phương trình Lagrange.
Lực dẫn động cho khớp tịnh tiến và mô men dẫn động
cho khớp quay được tính xấp xỉ dựa trên mô hình rô
bốt với các khâu giả thiết cứng tuyệt đối. Kết quả mô
phỏng việc phân tích động lực học ngược mô tả giá trị
lực/mô men dẫn động giữa mô hình cứng và mô hình
đàn hồi cùng với giá trị sai lệch giữa chúng. Giá trị
chuyển vị đàn hồi tại điểm thao tác cuối cũng được
thể hiện. Tuy nhiên, vẫn còn rất nhiều vấn đề cần
nghiên cứu thêm trong tương lai bởi giá trị sai lệch
của biến khớp trong thuật toán giải động lực học
ngược vẫn chưa được xét đến trong bài báo này.
Từ khóa - Động lực học ngược, khâu đàn hồi,
khớp tịnh tiến, chuyển vị đàn hồi
Dương Xuân Biên, Chu Anh Mỳ, Phan Bùi Khôi
Phân tích động lực học rô bốt có khâu đàn hồi
với các khớp tịnh tiến và khớp quay
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